如图所示，ABCD 是一个矩形，E 和 F 分别是 AD 和 BC 上的点。 三角形 ABF 的面积与矩形 ABCD 的面积之比为 1 : 9。 三角形 AFE 的面积与矩形 ABCD 的面积之比为 1 : 6。 请给出清晰的计算过程，并说明以下陈述是正确还是错误。

1. Write both triangle areas as fractions of the rectangle area.

Let the area of rectangle $ABCD$ be $R$.

Given
$\text{area of }\triangle ABF=\frac{1}{9}R$
and
$\text{area of }\triangle AFE=\frac{1}{6}R.$

2. Find and simplify the ratio of the two triangle areas.

$\text{area of }\triangle ABF : \text{area of }\triangle AFE$
$=\frac{1}{9}R : \frac{1}{6}R$
$=\frac{1}{9}:\frac{1}{6}$
Multiply both parts by $18$:
$\frac{1}{9}\times18 : \frac{1}{6}\times18 = 2:3.$

3. Compare with the statement and conclude true or false.

The ratio obtained is $2:3$, which is the same as the statement.

1. Express the given area ratio for triangle ABF using the rectangle dimensions.

Let the width of rectangle $ABCD$ be $w$ and the height be $h$.

Then
$\text{area of rectangle }ABCD = wh.$

Since $F$ lies on $BC$,
$\text{area of }\triangle ABF = \frac12 \times AB \times BF = \frac12 w(BF).$

Given
$\frac{\text{area of }\triangle ABF}{\text{area of rectangle }ABCD}=\frac19,$

$\frac{\frac12 w(BF)}{wh}=\frac19.$

2. Find the fraction of the rectangle height represented by BF.

Simplifying,
$\frac{BF}{2h}=\frac19.$

Hence
$\frac{BF}{h}=\frac29.$

3. Find the fraction of the rectangle height represented by FC.

Since $BC=h$ and $BF+FC=BC$,
$\frac{FC}{h}=1-\frac29=\frac79.$

4. Find the area ratio of triangle DFC to the rectangle and compare with the statement.

Now
$\text{area of }\triangle DFC=\frac12 \times DC \times FC=\frac12 w(FC).$

Therefore

So
$\text{area of }\triangle DFC : \text{area of rectangle }ABCD = 7:18.$

Since $7:18 \ne 2:5$, the statement is false.

1. Express the known triangle areas as fractions of the rectangle area.

Let the area of rectangle $ABCD$ be $R$.

From the question,
$\text{area}(\triangle ABF)=\frac{1}{9}R,$
$\text{area}(\triangle AFE)=\frac{1}{6}R.$

From part (B),
$\text{area}(\triangle DFC)=\frac{7}{18}R.$

2. Find the area of triangle EFD as the remaining part of the rectangle.

The four triangles $ABF$, $AFE$, $EFD$ and $DFC$ make up the rectangle.

$\text{area}(\triangle EFD)=R-\left(\frac{1}{9}R+\frac{1}{6}R+\frac{7}{18}R\right)$

Using denominator $18$,
$\text{area}(\triangle EFD)=R-\left(\frac{2}{18}R+\frac{3}{18}R+\frac{7}{18}R\right)$

$=R-\frac{12}{18}R=\frac{6}{18}R=\frac{1}{3}R.$

3. Compare the two sums of areas stated in the statement.

Left-hand side:
$\text{area}(\triangle ABF)+\text{area}(\triangle DFC)=\frac{1}{9}R+\frac{7}{18}R$
$=\frac{2}{18}R+\frac{7}{18}R=\frac{9}{18}R=\frac{1}{2}R.$

Right-hand side:
$\text{area}(\triangle AFE)+\text{area}(\triangle EFD)=\frac{1}{6}R+\frac{1}{3}R$
$=\frac{1}{6}R+\frac{2}{6}R=\frac{3}{6}R=\frac{1}{2}R.$

4. Conclude whether the statement is true or false.

Since both sums are equal to $\frac{1}{2}R$, the two sums of areas are equal.

Therefore, statement (C) is true.